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Example 5.1 Worked on the Board!PowerPoint Presentation

Example 5.1 Worked on the Board!

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Example 5.1 Worked on the Board!

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- Find the gravitational potential Φ inside & outside a spherical shell, inner radius b, outer radius a. (Like a similar electrostatic potential problem!)
- This is an important & fundamental problem in gravitational theory! If you understand this, you understand the gravitational potential concept!(& probably the electrostatic potential concept as well!) .
- Could approach the spherical shell using forces directly (Prob. 5.6), but its MUCH easier with the potential!

- Find Φ inside & outside a spherical shell of mass M, mass density ρ,inner radius b & outer radius a.
Φ = -G∫[ρ(r)dv/r]. Integrate over V.The difficulty, of course, is

properly setting up the integral!If properly set up, doing it is easy!

Recall Spherical Coordinates

Outline of Calculation!

Summary of ResultsM(4π)ρ(a3 - b3)

outside the shell

R > a, Φ = -(GM)/R (1)

The same as ifM were a point mass at the origin!

completely inside the shell

R < b, Φ = -2πρG(a2 - b2) (2)

Φ = constant, independent of position.

within the shell

b R a, Φ = -4πρG[a2- (b3/R) - R2] (3)

- Also, Φis continuous!
If R a, (1) & (3) are the same! If R b, (2) & (3) are the same!

- These results arevery important,especially
those forR > a, Φ = -(GM)/R

This says: The potential at any point outside

a spherically symmetric distribution of matter(shell or solid; a solid is composed of infinitesimally thick shells!)is independent of the size of the distribution & is the same as that for a point mass at the origin.

To calculate the potential for such a distribution we can consider all mass to be concentrated at the center.

- Also, the results are very important for
R < b, Φ = -2πρG(a2 - b2)

The potential is constant anywhere

inside a spherical shell.

The force on a test mass m

inside the shell is 0!

- Given the results for the potential Φ,we can compute the GRAVITATIONAL FIELD g inside, outside & within the spherical shell: g - Φ
- Φdepends on R only g is radially directed
g = g er = - (dΦ/dR)er [M(4π)ρ(a3 - b3)]

outside the shellR > a, g = - (GM)/R2The same as ifM were a point mass at the origin!

completely inside the shell R < b, g = 0

Since Φ = constant, independent of position.

within the shell

b R a, g = (4π)ρG[(b3/R2) - R]

- Plots of the
potential

Φ & the field g

inside, outside &

within a

spherical shell.

g - Φ

g = - (dΦ/dR)

Φ = constant

Φ = -(GM)/R

g = - (GM)/R2

g = 0